Problem Description
Andy the smart computer science student was attending an algorithms class when the professor asked the students a simple question, "Can you propose an efficient algorithm to find the length of the largest palindrome in a string?"
A string is said to be a palindrome if it reads the same both forwards and backwards, for example "madam" is a palindrome while "acm" is not.
The students recognized that this is a classical problem but couldn't come up with a solution better than iterating over all substrings and checking whether they are palindrome or not, obviously this algorithm is not efficient at all, after a while Andy raised his hand and said "Okay, I've a better algorithm" and before he starts to explain his idea he stopped for a moment and then said "Well, I've an even better algorithm!".
If you think you know Andy's final solution then prove it! Given a string of at most 1000000 characters find and print the length of the largest palindrome inside this string.
Input
Your program will be tested on at most 30 test cases, each test case is given as a string of at most 1000000 lowercase characters on a line by itself. The input is terminated by a line that starts with the string "END" (quotes for clarity).
Output
For each test case in the input print the test case number and the length of the largest palindrome.
Sample Input
abcbabcbabcba
abacacbaaaab
ENDSample Output
Case 1: 13
Case 2: 6
题意：给出多组数据，每组给出一个字符串，要求找出字符串中最长回文子串的长度，数据以 END 为结尾
思路：马拉车算法模版题
Source Program
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<string>
#include<cstring>
#include<cmath>
#include<ctime>
#include<algorithm>
#include<utility>
#include<stack>
#include<queue>
#include<vector>
#include<set>
#include<map>
#define PI acos(-1.0)
#define E 1e-9
#define INF 0x3f3f3f3f
#define N 1000001
#define LL long long
const int MOD=20091226;
const int dx[]= {-1,1,0,0};
const int dy[]= {0,0,-1,1};
using namespace std;
char str[N];//原字符串
char newStr[N*2];//预处理后的字符串
int p[N*2];//辅助数组
int init(){//对原字符进行预处理
newStr[0]='$';
newStr[1]='#';
int j=2;
int len=strlen(str);
for (int i=0;i<len;i++){
newStr[j++]=str[i];
newStr[j++]='#';
}
newStr[j] ='\0'; //字符串结束标记
return j;//返回newStr的长度
}
int manacher(){
int len=init();//取得新字符串长度并完成字符串的预处理
int res=-1;//最长回文长度
int id;
int mx=0;
for(int i=1;i<len;i++){
int j=2*id-i;//与i相对称的位置
if(i<mx)
p[i]=min(p[j], mx-i);
else
p[i]=1;
//由于左有'$',右有'\0'，不需边界判断
while(newStr[i-p[i]] == newStr[i+p[i]])//p[i]的扩大
p[i]++;
if(mx<i+p[i]){//由于希望mx尽可能的远，因此要不断进行比较更新
id=i;
mx=i+p[i];
}
res=max(res,p[i]-1);
}
return res;
}
int main(){
int Case=1;
while(scanf("%s",str)!=EOF){
if(str[0]=='E')
break;
printf("Case %d: %d\n",Case++,manacher());
}
return 0;
}